4 ms·
In fact, for b = 5, that infinite sum is equal to about 1.89. Substituting this into the upper and lower bounds for e(5) suggests that the expected numb
by whoopdedo 11y ago
In fact, for b = 5, that infinite sum is equal to about 1.89.
Substituting this into the upper and lower bounds for e(5)
suggests that the expected number of positive integers (greater than 1)
expressible with only “1” and “0” in bases 2 through to 5 should
lie between 0.08 and 0.59.
Of course I had to go on to b=6 and found the probability between 0.005 and 0.112
- mathistopheles 11y agoThe expectations are not independent though. There can be no more solutions for b=6 than there are for b=5. Given what we already know, we should calculate those sums, not from i=2, but from i=10^2184 or so, because we there are no solutions below that. The theoretical expected number of solutions for b=6 is therefore vanishingly small.